Graph this system of equations and solve. $y = \dfrac{3}{4} x + 1$ $-x-4y = 12$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Solution: The y-intercept for the first equation is $1$ , so the first line must pass through the point $(0, 1)$ The slope for the first equation is $\dfrac{3}{4}$ . Remember that the slope tells you rise over run. So in this case for every $3$ positions you move up You must also move $4$ positions to the right. $4$ positions to the right. $3$ positions up from $(0, 1)$ is $(4, 4)$ Graph the blue line so it passes through $(0, 1)$ and $(4, 4)$ Convert the second equation, $-x-4y = 12$ , to slope-intercept form. $y = -\dfrac{1}{4} x - 3$ The y-intercept for the second equation is $-3$ , so the second line must pass through the point $(0, -3)$ The slope for the second equation is $-\dfrac{1}{4}$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move down (because it's negative) You must also move $4$ positions to the right. $4$ positions to the right. Graph the green line so it passes through $(0, -3)$ and $(4, -4)$ The solution is the point where the two lines intersect. The lines intersect at $(-4, -2)$.